tugas_4.docx

# tugas_4.docx - a Gambar Diagrammomen primer dan diagram...

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a) Gambar Diagrammomen primer dan diagram momen sekunder 1. Buat persamaan parabola bentang AB y = A x 2 + Bx + C masukan 3 titik yang diketahui x y 0 200 12 -400 24 200 diperoleh persamaan y = 25 6 x 2 100 x + 200 dy dx = 25 3 x 100 = θ untuk x =0 1 =0.1 P AB =Pe* 1 =500 kN P BA = PAB = 500 kN w p 1 = 8 P ( e 1 + e 2 ) L 2 = 8 5000 0.6 24 2 = 125 3 kN m

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2. Persamaan parabola bentang BC masukan 3 titik yang diketahui x y 0 200 12 -300 24 0 diperoleh persamaan y = 25 9 x 2 75 x + 200 dy dx = 50 9 x 75 = θ untuk x =0 3 =0.075 untuk x=24 4 =175/3 P BC =Pe* 3 =0.075*5000=375 kN P CB =Pe* 4 =(175/3000)*5000=875/3 kN w p 1 = 8 P ( e 1 + e 2 ) L 2 = 8 5000 0.4 24 2 = 250 9 kN m untuk penyederhanaan titik maksimum di tengah bentang , sehingga e1=(200+0)/2= 100 mm P A = P AB =500 kN P B = P BA +P BC = 875 kN P C =P CB =291.67 kN semua dalam arah downward , dan w ps dalam arah upward 3. Analisis momen sekunder dengan FEM Keterangan : A B C AB BA BC CB
Stiffness Coefficient - 4 EI L 4 EI L - Distribution Factor 0 0.5 0.5 0 FEM 1000 -3000.24 2000.16 0 CO 0 500 0 0 BAL 0 250.04 250.04

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